矩阵快速幂（hdu5171GTY's birthday gift）

最新推荐文章于 2019-02-18 22:58:22 发布

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最新推荐文章于 2019-02-18 22:58:22 发布

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分类专栏：矩阵快速幂

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本文介绍了一个关于为FFZ准备生日礼物的有趣问题，GTY决定制作一个多集来作为礼物，通过使用特定的魔法操作使多集的总和最大化。文章详细解释了解决方案，包括如何通过构造矩阵并利用快速幂运算求解最大和。

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GTY's birthday gift

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 307 Accepted Submission(s): 111

Problem Description

FFZ's birthday is coming. GTY wants to give a gift to ZZF. He asked his gay friends what he should give to ZZF. One of them said, 'Nothing is more interesting than a number multiset.' So GTY decided to make a multiset for ZZF. Multiset can contain elements with same values. Because GTY wants to finish the gift as soon as possible, he will use JURUO magic. It allows him to choose two numbers a and b(

a,b∈S ), and add

a+b to the multiset. GTY can use the magic for k times, and he wants the sum of the multiset is maximum, because the larger the sum is, the happier FFZ will be. You need to help him calculate the maximum sum of the multiset.

Input

Multi test cases (about 3) . The first line contains two integers n and k (

2≤n≤100000,1≤k≤1000000000 ). The second line contains n elements

ai (

1≤ai≤100000 )separated by spaces , indicating the multiset S .

Output

For each case , print the maximum sum of the multiset (

mod 10000007 ).

Sample Input

  
   3 2
3 6 2

Sample Output

Source

BestCoder Round #29

每次挑选两个最大的进行加和，然后放到集合里

构造矩阵

| 0 1 0 |

A=| 1 1 0 |

| 1 1 1 |

| a |

B=| b |

| a+b |

其中a代表集合中第二大，b代表最大

A^k*B的最后一项加上集合中除去a，b其他元素的和就是答案

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn=100010;
const int maxm=5;
const int MOD=10000007;
int N,a[maxn],K;
struct Matrix
{
    LL mat[maxm][maxm];
    Matrix(){memset(mat,0,sizeof(mat));}
    Matrix operator * (Matrix A)
    {
        Matrix res;
        for(int i=0;i<3;i++)
        {
            for(int j=0;j<3;j++)
            {
                for(int k=0;k<3;k++)
                    res.mat[i][j]=(res.mat[i][j]+(mat[i][k]*A.mat[k][j])%MOD)%MOD;
            }
        }
        return res;
    }
};
Matrix pow_mul(Matrix A,int n)
{
    Matrix res;
    for(int i=0;i<maxm;i++)res.mat[i][i]=1;
    while(n)
    {
        if(n&1)res=res*A;
        A=A*A;
        n>>=1;
    }
    return res;
}
int main()
{
    while(scanf("%d%d",&N,&K)!=EOF)
    {
        for(int i=1;i<=N;i++)scanf("%d",&a[i]);
        sort(a+1,a+1+N);
        Matrix A,B;
        A.mat[0][1]=A.mat[1][0]=A.mat[1][1]=A.mat[2][0]=A.mat[2][1]=A.mat[2][2]=1;
        A=pow_mul(A,K);
        B.mat[0][0]=a[N-1],B.mat[1][0]=a[N],B.mat[2][0]=a[N-1]+a[N];
        A=A*B;
        LL sum=0;
        for(int i=1;i<N-1;i++)sum+=a[i];
        sum=(sum+A.mat[2][0])%MOD;
        printf("%I64d\n",sum);
    }
    return 0;
}